Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 2 - Section 2.3 - Writing Equations of Lines - 2.3 Exercises: 43

Answer

$x-2y=-13$

Work Step by Step

Using $y-y_1=m(x-x_1)$ or the point-slope form of linear equations, the equation of the line passing through $( -5,4 )$ and with a slope of $m= \dfrac{1}{2} $ is \begin{array}{l}\require{cancel} y-4=\dfrac{1}{2}(x-(-5)) \\\\ y-4=\dfrac{1}{2}(x+5) .\end{array} In the form $y=mx+b$, the equation above is equivalent to \begin{array}{l}\require{cancel} y-4=\dfrac{1}{2}(x+5) \\\\ y-4=\dfrac{1}{2}x+\dfrac{5}{2} \\\\ y=\dfrac{1}{2}x+\dfrac{5}{2}+4 \\\\ y=\dfrac{1}{2}x+\dfrac{5}{2}+\dfrac{8}{2} \\\\ y=\dfrac{1}{2}x+\dfrac{13}{2} .\end{array} In the form $Ax+By=C$, the equation above is equivalent to \begin{array}{l}\require{cancel} y=\dfrac{1}{2}x+\dfrac{13}{2} \\\\ 2(y)=2\left( \dfrac{1}{2}x+\dfrac{13}{2} \right) \\\\ 2y=x+13 \\\\ -x+2y=13 \\\\ x-2y=-13 .\end{array} Hence, the different forms of the equation of the line with the given conditions are \begin{array}{l}\require{cancel} \text{slope-intercept form: } y=\dfrac{1}{2}x+\dfrac{13}{2} \\\\ \text{standard form: } x-2y=-13 .\end{array}
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